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Anastasy [175]
4 years ago
9

Name 3 constellations that are on the ecliptic and visible at this time

Physics
1 answer:
Hoochie [10]4 years ago
6 0

Answer:

also right now we are in ZODIAC SEASON OF  Libra WE JUST PASSED VIRGO AND WE ARE HEADING INTO the constellations of fall  are Aquarius, Grus, Lacerta, Octans, Pegasus and Piscis Austrinus. Lacerta and Pegasus are located in the northern sky

Explanation:

The five northern constellations visible from most locations north of the equator throughout the year are Cassiopeia, Cepheus, Draco, Ursa Major, and Ursa Minor.

The three southern circumpolar constellations visible from most locations in the southern hemisphere are Carina, Centaurus, and Crux.

The ecliptic currently passes through the following constellations:

Pisces.

Aries.

Taurus.

Gemini.

Cancer.

Leo.

Virgo.

Libra.    hope it helps

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Answer:

W of the person in moon  ≈ 124.70 N

Explanation:

Weight: Weight of a body can be defined as the product of mass and the gravitational acceleration of the body. The S.I unit of weight is  Newton (N). It can be expressed mathematically as

W = mg

Where W = weight of the body, m = mass of the body (kg) and a = acceleration of the body (m/s²)

Weight(W) = Mass (m) × Acceleration due to gravity (g)

∴ W = m × g.

If the person is on the moon,

Mass = 76.5 kg.

g (moon) = 16.6% of g ( earth)

But g(earth) = 9.80 m/s².

∴ g (moon) = 9.80 × (16.6/100)

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
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Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

Now, this force is no other than centripetal force, so we can write this equation:

Ff  = Fc ⇒ μ*m*g = m*ac

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μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each  other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

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Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

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